Outlet scaling with diameter (area proportional to d^2) A tap of diameter d empties a tank in 40 minutes. Assuming the discharge is proportional to the cross-sectional area, how long will another tap of diameter 2d take to empty the same tank?

Introduction / Context:Under similar head/pressure conditions, volumetric discharge of a round tap is proportional to cross-sectional area, which scales with the square of the diameter (d^2). Doubling diameter multiplies area (and hence flow) by 4, reducing time by a factor of 4.

Given Data / Assumptions:

• Tap 1 diameter = d; time = 40 min.
• Tap 2 diameter = 2d; flow ∝ (2d)^2 = 4 d^2.
• Same tank, comparable hydraulic conditions; time inversely ∝ area.

Concept / Approach:If Q ∝ d^2 and V is fixed, then time t ∝ 1/Q ∝ 1/d^2. Doubling diameter quarters the time.

Step-by-Step Solution:

Verification / Alternative check:Relative-flow logic: 4× the flow in the larger tap empties the same volume in one-quarter the time.

Why Other Options Are Wrong:20 min halves, not quarters, the time; 5 min would imply 8× flow; 80 min is the opposite trend.

Common Pitfalls:Scaling time linearly with diameter instead of with area (d^2).

Final Answer:10 min